Question

4 cm tall object is placed
perpendicular to the principal axis of
a convex lens of focal length 24 cm.
The distance of the object from the
lens is 16 cm. Find the position, size
and nature of the image formed,
using the lens formula. [ *
O V=-28 cm and h=+12 cm
O V=-18 cm and h=+12 cm
O V=-48 cm and h=+12 cm
O V=-48 cm and h= 2 cm
O V=-48 cm and h=+1.2 cm
O V=-480 cm and h=+12 cm​

Answer 1

Given :

In convex lens,

Object height : 4 cm

focal length : 24 cm

Object distance : 16 cm

To find :

The position, size and nature of the image formed.

Solution :

Using lens formula that is,

» The formula which gives the relationship between image distance, object distance and focal length of a lens is known as the lens formula.

The lens formula can be written as :

$\boxed{ \bf \dfrac{1}{v} - \dfrac{1}{u}= \dfrac{1}{f}}$

where,

• v denotes image distance
• u denotes object distance
• f denotes focal length

by substituting all the given values in the formula,

$\dashrightarrow \sf \dfrac{1}{v} - \dfrac{1}{u}= \dfrac{1}{f}$

$\dashrightarrow \sf \dfrac{1}{v} - \dfrac{1}{-16}= \dfrac{1}{24}$

$\dashrightarrow \sf \dfrac{1}{v} + \dfrac{1}{16}= \dfrac{1}{24}$

$\dashrightarrow \sf \dfrac{1}{v} = \dfrac{1}{24}- \dfrac{1}{16}$

$\dashrightarrow \sf \dfrac{1}{v} = \dfrac{2 - 3}{48}$

$\dashrightarrow \sf \dfrac{1}{v} = \dfrac{- 1}{48}$

$\dashrightarrow \sf v = -48 cm$

Thus, the focal length of the image is -48 cm.

we know that,

» The ratio of image distance to the object distance is equal to the the ratio of image height to the object height

$\dashrightarrow \sf \dfrac{h'}{h} = \dfrac{v}{u}$

$\dashrightarrow \sf \dfrac{h'}{4} = \dfrac{-48}{-16}$

$\dashrightarrow \sf \dfrac{h'}{4} = 3$

$\dashrightarrow \sf h' = 12 cm$

Thus, the height of the image is 12 cm.

Nature of the image :

• The image is real and inverted.
• The image is highly enlarged.

Answer 2

Bonjour ⚘⚘

$Given :-$

• Object height = 4 cm
• focal length = 24 cm
• Object distance = 16 cm

$To\: find :-$

• Position of the image
• Size of the image
• Nature of the image

$Solution :-$

By using lens formula,

$:\implies\:\: \dfrac{1}{v} - \dfrac{1}{u}= \dfrac{1}{f}$

Where,

• v = image distance
• u = object distance
• f = focal length

$:\implies\:\: \dfrac{1}{v} - \dfrac{1}{u}= \dfrac{1}{f}$

$:\implies\:\: \dfrac{1}{v} - \dfrac{1}{-16}= \dfrac{1}{24}$

$:\implies\:\: \dfrac{1}{v} + \dfrac{1}{16}= \dfrac{1}{24}$

$:\implies\:\: \dfrac{1}{v} = \dfrac{1}{24}- \dfrac{1}{16}$

$:\implies\:\: \dfrac{1}{v} = \dfrac{2 - 3}{48}$

$:\implies\:\: \dfrac{1}{v} = \dfrac{- 1}{48}$

$:\implies\:\: v = -48 cm$

• Focal length of the image is -48 cm.

Now, finding the height of the image,

$:\implies\:\: \dfrac{h'}{h} = \dfrac{v}{u}$

$:\implies\:\: \dfrac{h'}{4} = \dfrac{-48}{-16}$

$:\implies\:\: \dfrac{h'}{4} = 3$

$:\implies\:\: h' = 12 cm$

Neight of the image is 12 cm.

Nature of the image,

• Real, inverted and highly enlarged.